Center of Mass and Linear Momentum - PowerPoint Presentation

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Center of Mass and Linear Momentum - PPT Presentation

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Slide1

Center of Mass and Linear Momentum

Chapter 9

9-1 Center of Mass

Motion of rotating objects is complicated

There is a special point for which motion is simple

Center

of mass

of bat

traces out a parabola, just as a tossed ball

doesAll other points rotate around this point

Figure 9-1

Goals for Chapter 9To learn the meaning of the momentum of a particle and how an impulse causes it to changeTo learn how to use the principle of conservation of momentumTo learn how to solve problems involving collisionsSlide4

Goals for Chapter 8To learn the definition of the center of mass of a system and what determines how it movesTo analyze situations, such as rocket propulsion, in which the mass of a moving body changesSlide5

IntroductionIn many situations, such as a bullet hitting a carrot, we cannot use Newton’s second law to solve problems because we know very little about the complicated forces involved.In this chapter, we shall introduce momentum and impulse, and the conservation of momentum, to solve such problems.Slide6

9-1 Center of Mass

DEFINITION:

center of mass (com) of a system of particles:

For two particles separated by a distance d, where the origin is chosen at the position of particle 1:

9-1

Center of Mass

For two particles separated by a distance d, where the origin is chosen at the position of particle 1:

For two particles, for an arbitrary choice of origin:

9-1

Center of Mass

9-1 Center of Mass

For many particles, we can generalize the equation, where M = m1 + m

2 + . . . + mn:

9-1 Center of Mass

In three dimensions, we find the center of mass along each axis separately:

More concisely, we can write in terms of vectors:Slide11

9-1 Center of Mass

For solid bodies, we take the limit of an infinite sum of infinitely small particles → integration!

Coordinate-by-coordinate, we write:

Here

is the mass of the object

9-1 Center of Mass

If objects have uniform density = ρ (“rho”)

Substituting, we find the center of mass simplifies:

9-1 Center of Mass

The center of mass lies at a point of symmetry

It lies on the line or plane of symmetry)

It need not be on the object (consider a doughnut)

9-1 Center of Mass

Answer: (a

) at the origin (b) in Q4, along y=

-x

(d) at the origin (e) in Q3, along y=x (f) at the origin

9-1 Center of Mass

Example Subtracting

Task: find COM of a disk with another disk taken out of it:

Find

the

COM

of the two individual COMs (one for each disk), treating the cutout as having negative

9-1 Center of Mass

Example Subtracting

On the diagram, comC is the center of mass for Plate P and Disk S combined

com

is the center of mass for the composite plate with Disk

9-2 Newton's Second Law for a System of Particles

Center of mass motion continues unaffected by forces internal to a system (collisions between billiard balls)

Motion of a system's center of mass:

Reminders:

net

is the sum of all

external

forces

is the total, constant, mass of the

closed

system

com

is the

center of mass

acceleration

6-2 Newton's Second Law for a System of Particles

Examples Using the center

of mass motion equation:Billiard collision: forces are only internal,

F = 0 so a

6-2 Newton's Second Law for a System of Particles

Examples Using the center

of mass motion equation:Billiard collision: forces are only internal,

F = 0 so a

= 0

Baseball bat:

6-2 Newton's Second Law for a System of Particles

Examples Using the

center of mass motion equation:

Exploding rocket: explosion forces are internal, so only the gravitational force acts on the system, and the

COM

follows a gravitational trajectory

9-2 Newton's Second Law for a System of Particles

Answer: The system consists of Fred, Ethel and the pole. All forces are internal. Therefore the com will remain in the same place. Since

the origin is the com, they will meet at the origin in all three cases! (Of course the origin where the com is located is closer to Fred than to Ethel.)

Momentum and Newton’s second lawThe momentum of a particle is the product of its mass and its velocity:

What is the momentum of a 1000kg car going

25 m/s west?Slide26

Momentum and Newton’s second lawThe momentum of a particle is the product of its mass and its velocity:

What is the momentum of a 1000kg car going

25 m/s west?

= m

= (1000 kg)(25 m/s) = 25,000

kgm/s westSlide27

9-3 Linear Momentum

Momentum

Points in the same direction as the velocity

Can only be changed by a net external force

We can write Newton's second law thus:

9-3 Linear Momentum

We can write Newton's second law thus:

9-3 Linear Momentum

Answer: (a) 1, 3, 2 & 4 (b) region 3

We can sum momenta for a system of particles to find:Slide32

9-3 Linear Momentum

Taking time derivative write Newton's second law for system of particles as

Net external

force

system changes linear momentum

Without a net external force, the total linear momentum of a system of particles cannot change

This is called the

law of conservation of linear momentum

9-5

Conservation of Linear Momentum

Impulse and momentumImpulse of a force is product of force & time interval during which it acts.Impulse is a vector!On a graph of Fx

versus time, impulse equals area under curve.Slide35

Impulse and momentumImpulse-momentum theorem: Impulse = Change in momentum J of particle during time interval equals net force acting on particle during intervalJ = Dp =

pfinal – pinitial J = Net Force x time = (

SF) x (Dt)so…Dp = (

SF) x (

) SF = Dp/(D

t)Note!! J , p, Fare all VECTORS!)Slide36

9-4 Collision and Impulse

If F isn’t constant over time….

This means that the applied impulse

is equal to the change in momentum of the object during the collision:

9-4 Collision and Impulse

Given Favg and duration:

We are integrating: we only need to know the area under the force curve

9-4 Collision and Impulse

Answer:

(a) zero (b) positive (c) along the positive

-axis (normal force)

For an impulse of zero we find:

Which is another way to say momentum is conserved!

Law

of conservation of linear

momentum

9-5

Conservation of Linear Momentum

9-5

Conservation of Linear Momentum

Check components of

net external force to determine if you should apply

conservation of momentum

9-5 Conservation of Linear Momentum

Internal forces can change momenta of parts of the system, but cannot change the linear momentum of the entire system

9-5 Conservation of Linear Momentum

Internal forces can change momenta of parts of the system, but cannot change the linear momentum of the entire system

Answer:

(a) zero (b) no (c) the negative

direction

9-5 Conservation of Linear Momentum

Do not confuse momentum and energy

Change in KE => SPEED changes (+ or – directions)

Change in P

=> direction may have changed, or

Compare momentum and kinetic energyChanges in momentum depend on time over which net force acts But… Changes in kinetic energy depend on the distance over which net force acts. Slide48

Ice boats againTwo iceboats race on a frictionless lake; one with mass m & one with mass 2m.Wind exerts same force on both.Both boats start from rest, both travel same distance to finish line.Questions!Which crosses finish line with

more KE?Which crosses with more Momentum?Slide49

Ice boats again Two iceboats race on a frictionless lake; one with mass m and one with mass 2m.The wind exerts the same force on both.Both boats start from rest, and both travel the same distance to the finish line?Which crosses the finish line with more KE?

In terms of work done by wind?

W =

KE = Force x distance = same

; So ½ m1v1

2 = DKE1 = W = DKE2 = ½ m2v22 Slide50

In terms of work done by wind?

W =

KE = Force x distance = same;

So ½ m1v12 =

DKE1 = W = DKE2 = ½ m2v22 Since m2 > m1, v2 < v1 so more massive boat losesSlide51

In terms of momentum?

Force on the boats is the same for each, but

TIME

that force acts is different. The second boat accelerates slower, and takes a longer time.

Since t

2> t1, p2 > p1 so second boat has more momentum Slide52

Ice boats again Check with equations!Force of wind = same; distance = same Work done on boats is the same, gain in KE same.½ m1v12 = DKE1 = W = D

KE2 = ½ m2v22And m1v

1 = p1; m2v2 = p2½ m

1v12 =

½ (m

v1) v1 =

½ p1 v1 & same for ½ p2 v2½ p1 v1 = ½ p2 v2Since V1 > V2, P2 must be greater than P1!p2 > p1Slide53

A ball hits a wall A 0.40 kg ball moves at 30 m/s to the left, then rebounds at 20 m/s to the right from a wall. Slide54

A ball hits a wall A 0.40 kg ball moves at 30 m/s to the left, then rebounds at 20 m/s to the right from a wall. What is the impulse of net force during the collision, and if it is in contact for 0.01 s, what is the average force acting from the wall on the ball?Slide55

Remember that momentum is a vector!When applying conservation of momentum, remember that momentum is a vector quantity!Slide56

Remember that momentum is a vector!When applying conservation of momentum, remember that momentum is a vector quantity!Use vector addition to add momenta in COMPONENTS!Slide57

Kicking a soccer ball – Example 8.3Soccer ball 0.40 kg moving left at 20 m/s, then kicked up & to the right at 30 m/s at 45 degrees. If collision time is 0.01 seconds, what is impulse?Slide58

9-6 Types of Collisions

Elastic collisions:

Total kinetic energy is unchanged (conserved)

A useful approximation for common situations

In real collisions, some energy is always

transferred

Inelastic collisions

: some energy is transferredCompletely inelastic collisions:The objects stick togetherGreatest loss of kinetic energy

For one dimension inelastic collision

9-6

Momentum and Kinetic Energy in Collisions

Completely inelastic collision, for target at rest:

9-6

Momentum and Kinetic Energy in Collisions

9-6 Momentum and Kinetic Energy in Collisions

9-7 Elastic Collisions in One Dimension

Total kinetic energy is conserved in elastic collisions

9-7 Elastic Collisions in One Dimension

For a stationary target, conservation laws give:

Figure 9-18

9-7 Elastic Collisions in One Dimension

With some algebra we get:

Results

Equal masses:

= 0, v2f = v1i: the first object stops

Massive target, m2 >> m1: the first object just bounces back, speed mostly unchangedMassive projectile: v1f ≈ v1i, v2f ≈ 2v1i: the first object keeps going, the target flies forward at about twice its speed

9-7 Elastic Collisions in One Dimension

For a target that is also moving, we get:

9-8 Collisions in Two Dimensions

Apply conservation of momentum along each axis

Apply conservation of energy for elastic collisions

Example

For

stationary target:

Along x:Along y:Energy:

A two-dimensional collisionTwo robots collide and go off at different angles.You must break momenta into x & y components and deal with each direction separatelySlide69

9-8 Collisions in Two Dimensions

Answer:

(a) 2 kg m/s (b) 3 kg m/s Slide71

Elastic collisionsIn elastic collision, total momentum of system in a direction is same after collision

as before… if no external forces act in that direction:Pix = Pfx

mavai +mb

vbi =

vaf +mbv

bfBut wait – there’s more! Slide72

Objects colliding along a straight lineTwo gliders collide on an frictionless track.What are changes in velocity and momenta?Slide73

Elastic collisionsIn an elastic collision, the total kinetic energy of the system is the same after the collision as before.KEi = KEf

½ mavai2 + ½ mbv

bi2 = ½ mavaf2 + ½ m

bvbf

= Slide74

Elastic collisionsIn an elastic collision, Difference in velocities initially = (-) difference in velocities finally(vai - vbi) = - (v

af – vbf)Slide75

Elastic collisionsIn an elastic collision, Difference in velocities initially = (-) difference in velocities finally(vai - vbi) = - (v

af – vbf)Solve generally for final velocities:vaf = (m

a-mb)/(ma+mb

)vai + 2m

ma+mb)v

bivbf = (mb-ma)/(ma+mb)vbi + 2ma/(ma+mb)vaiSlide76

Elastic collisions Slide77

Elastic collisionsBehavior of colliding objects is greatly affected by relative masses.Slide78

Elastic collisions Slide79

Elastic collisionsBehavior of colliding objects is greatly affected by relative masses.Slide80

Elastic collisions Slide81

Elastic collisionsBehavior of colliding objects is greatly affected by relative masses.

Stationary Bowling Ball!Slide82

Inelastic collisions

any

collision where external

forces can be neglected,

total

momentum conserved.C

ollision when bodies stick together is completely inelastic collision In inelastic collision, total kinetic energy after collision is less than before collision.Slide83

Some inelastic collisionsCars are intended to have inelastic collisions so the car absorbs as much energy as possible.Slide84

The ballistic pendulumBallistic pendulums are used to measure bullet speedsSlide85

A 2-dimensional automobile collision Two cars traveling at right angles collide.Slide86

An elastic straight-line collisionSlide87

Neutron collisions in a nuclear reactor Slide88

A two-dimensional elastic collisionSlide89

Rocket propulsionConservation of momentum holds for rockets, too! F = dp/dt = d(mv

)/dt = m(dv/dt) + v(

dm/dt)As a rocket burns fuel, its mass decreases (dm< 0!)Slide90

Rocket propulsionInitial Values

= initial x-momentum of rocket

exh

= exhaust velocity from rocket motor

(This will be constant!) (v - vexh) = relative velocity of exhaust gases

m = initial mass of rocketv = initial velocity of rocket in x directionSlide91

Rocket propulsionChanging values

decrease

in mass of rocket from fuel

= increase in velocity of rocket in x-directiondt = time interval over which dm and dt changeSlide92

Rocket propulsionFinal values

m + dm

final

mass of rocket less fuel ejected

v + dv

= increase in velocity of rocket (m + dm) (v + dv) = final x-momentum of rocket (+x direction)

[- dm] (v - vexh) = final x-momentum of fuel (+x direction)Slide93

Rocket propulsionFinal values

mv = [(m + dm)

(v + dv)] +

[- dm]

(v - vexh)

initial momentum

final momentum of the lighter rocket

final momentum of the ejected mass of gasSlide94

Rocket propulsionFinal values

mv = [(m + dm)

(v + dv)] +

[- dm]

(v - vexh) mv = mv + mdv + vdm + dmdv –dmv +dmvexhSlide95

Rocket propulsionFinal values

mv = [(m +

)

(v + dv)

] + [- dm] (v - vexh) mv

= mv + mdv + vdm + dmdv –dmv +dmvexh mv = mv + mdv + vdm + dmdv –dmv +dmvexhSlide96

Rocket propulsionFinal values

= mv + mdv + vdm + dmdv –dmv +dmv

exhSlide97

Rocket propulsionFinal values

= mdv +

dmdv

dmv

exhSlide98

Rocket propulsionFinal values

0 = mdv +

dmdv

dmv

exh

Neglect the assumed small term:m(dv) = -dmdv – (dm)vexhSlide99

Rocket propulsionFinal values

m(dv) =

– (dm)v

exh

Gain in momentum of original rocket is related to rate of mass loss and exhaust velocity of gas!Slide100

Rocket propulsionFinal values

m(dv) =

– (dm)v

exh

Gain in momentum of original rocket is related to rate of mass loss and exhaust velocity of gas!

Differentiate both sides with respect to time!Slide101

Rocket propulsionFinal values

m(dv) =

– (

)

exhm(dv/dt) = – [(dm)/dt] vexhSlide102

Rocket propulsionFinal values

m(dv) =

– (dm)v

exh

m(dv/dt) =

– [(dm)/dt] v

exhma = Force (Thrust!) = – [(dm)/dt] (vexh)Slide103

Rocket propulsionFinal values

m(dv) =

– (

)

exhm(dv/dt) = – [(dm)/dt] vexh

ma = Force (Thrust!) = – [(dm)/dt] (vexh)rate of change of mass

Velocity of gas exhaustedSlide104

Rocket propulsion

Thrust = – [(dm)/dt] (v

exh)

Example:

exhaust

= 1600 m/s

Mass loss rate = 50 grams/secondThrust =?Slide105

Rocket propulsionFinal values

Thrust = – [(dm)/dt] (v

exh

)

Example

exhaust

= 1600 m/sMass loss rate = 50 grams/secondThrust = -1600 m/s (-0.05 kg/1 sec) = +80NSlide106

Rocket propulsionGain in speed?

m(dv) =

– (dm)v

exh

dv =

– [(dm)/m] v

exh integrate both sidesSlide107

Rocket propulsionGain in speed?

m(dv) =

– (

dm)

exh

dv = – [(dm)/m] vexh (integrate both sides)vf - v

i = (vexh) ln(m0/m) m0 = initial massm0 > m, so ln > 1“mass ratio”

Velocity of gas exhausted

gain in velocitySlide108

Rocket propulsion – Gain in Speed

- v

= (

vexh) ln(m0/m)

m0 = initial massm0 > m, so ln > 1“mass ratio”

Velocity of gas exhausted

gain in velocity

Faster exhaust, and greater difference in mass, means a greater increase in speed! Slide109

Rocket propulsionGain in speed?

- vi = (v

exh

) ln(m

0/m) Example:Rocket ejects gas at relative 2000 m/s. What fraction of initial mass is not fuel if final speed is 3000 m/s?Slide110

Rocket propulsionGain in speed?

- vi = (v

exh

) ln(m

0/m) Example:Rocket ejects gas at relative 2000 m/s. What fraction of initial mass is not fuel if final speed is 3000 m/s?ln(mo/m) = (3000/2000) = 1.5 so m/mo = e-1.5 = .223Slide111

Rocket propulsionGain in speed?

- vi = (v

exh

) ln(m

0/m)

m0 = initial massm0 > m, so ln > 1“mass ratio”

Velocity of gas exhausted

gain in velocity

STAGE rockets to throw away mass as they use up fuel, so that m

/m is even higher!